• Author / Uploaded
• Ari Wibowo Nugroho

• Categories
• Pandeo
• Viga (Estructura)
• Doblar
• Resistencia de materiales
• Ingeniería de productos químicos

Citation preview

INDEX 1.

SUMMARY ......................................................................................... 3

2.

INTRODUCTION ................................................................................ 4

3.

ASSIGNMENT .................................................................................... 5

INTRODUCTION................................................................................................. 5 4.

MODEL............................................................................................... 6

INTRODUCTION................................................................................................. 6 PLATE ............................................................................................................ 6

Deflection.................................................................................................. 7 Stress........................................................................................................ 7 Maximum Plate dimensions......................................................................... 7 Distance Between Profiles........................................................................... 8 PROFILE ......................................................................................................... 9

Required inertia ......................................................................................... 9 Profiles.................................................................................................... 10 Wall ........................................................................................................ 10 Neutral axis wall + profile......................................................................... 10 Total Inertia ............................................................................................ 11 Stress...................................................................................................... 11 BUCKLING ..................................................................................................... 12 WELDING ...................................................................................................... 13 FRAME ......................................................................................................... 14

Required Moments ................................................................................... 14 Actual moments of frame ......................................................................... 14 BASE ........................................................................................................... 15

Flat base ................................................................................................. 15 Skid Base ................................................................................................ 15 TOTAL TANK .................................................................................................. 15

Area........................................................................................................ 15 Weight .................................................................................................... 16 Total Inertia: ........................................................................................... 16 Lifting Lugs.............................................................................................. 16 Computer program................................................................................... 18 5.

ANALYSE ......................................................................................... 19

U PROFILES ................................................................................................... 19 TOTAL INERTIA............................................................................................... 20

Plate Stress ............................................................................................. 21 MAXIMUM DISTANCE BETWEEN PROFILES .............................................................. 22 SEISMIC ANALYSE............................................................................................ 23 6.

CONCLUSIONS ................................................................................ 24

7.

RECOMMENDATIONS ...................................................................... 25

8.

APPENDICES ................................................................................... 27

CLASSICAL PLATE EQUATION .............................................................................. 28

Flowchart main program........................................................................... 33 Flowchart function solution ....................................................................... 36 9.

APPENDIX PROFILES ...................................................................... 38

H PROFILES ................................................................................................... 38 I PROFILES .................................................................................................... 38 L PROFILES ................................................................................................... 38 BEND U PROFILES ........................................................................................... 40

1. Summary

2. Introduction

4. Model

3. Assignment

Introduction

Introduction The assignment coexists of a formulation of alternative strength and stiffness calculations on the tank wall of a transformer. With these calculations an optimalisation has to be made of the type and the number of reinforcements used on the wall of the tank. Pauwels has several programs and documents about the calculations, but in practice the tanks are too heavy in comparison with competitors. A next point of consideration is standardisation of the design. Each transformer will be designed according the desires and demands of the customers. At this moment there are too many variables in design, which means that the variation in the design is great. At this moment there are horizontal, vertical and a combination of horizontal and vertical reinforcements used Is it possible to reduce a certain number of variables in design, e.g. the same design for the lifting lugs for all transformers. The desires and demands of customers are changing to the direction of vertical reinforcements instead of horizontal because of reduction of corrosion on the outside wall.

The question is if and how the weight of a transformer tank can be reduced. Reduction of lead time for design Kill combination hv Optimalisation After an analyse of the tank, the calculations of the tank are too complex to make a simple and reliable model of it. The effects of the four walls on each other and the effects of the reinforcements connected to the bottom and top makes all models, also based on elasticity theory, only an approach of reality. When an optimalisation of the type and the number of reinforcements has to be made, the calculations have to be simple for checking all possibilities. That’s why there will be made a model of the walls separately with beam and plate theories. The output of these calculations will be a very good input for optimalisation with

The walls of the transformer tank are made of steel with thickness between 6 and 12 mm. The tank has to be designed to withstand full vacuum pressure. This vacuum is applied before filling the transformer with oil. The wall of the tank can be stiffened with reinforcements to reduce the displacement and the stresses in the plate. The reinforcements are steel H,I L,U profiles and welded to the tank wall according to figure xx Another critical load is lifting the transformer filled with the oil and the core. For the design of the transformer tank the “worst case scenario” has to be chosen. The restraints for the design are the maximum displacement and stress in the tanks walls. The assembled transformer tank is too complex for simple calculations for an optimalisation of the weight, therefore the tank is modelled as four walls with reinforcements and a Figure 1 respectively U, L, H and bottom plate, each wall is considered sepaI rately. The wall is split into sections where every section represents a reinforcement and 2 steel plates on both sides of the profile. The dimensions of these plates are half of the distance between the profiles and the height of the tank. These sections are checked for inertia and stress, the plates are checked for displacement and stress and finally the combination of the walls are checked for welding in the corners and buckling of the profiles. The model of the transformer tank is split in six sections where the different aspects of that part will be highlighted in order of: − Plate − Profile − Buckling − Welding − Frame − Base Figure 2 Clamped plate with uniform load

Plate ement Finite Element Method (FEM) in a Finite El Programme. Only with FEM a reliable presentation of reality can be given and the effects of all the corners, wall and reinforcements on each other.

Deflection The deflection f of a clamped plate can be estimated with α as dimensionless steel factor. The calculated displacement has to be smaller than the deflection factor times the smallest length of the plate. In this case it will be the distance between the profiles (DBP). The Distance Between Profiles will be discussed in the next paragraphs. p * ( DBP) 4 0.036 f =α * ;α = DBP 2 E * Tw3 1+ ( ) Hw f max = defl max * DBP

Stress The stress σ of a clamped plate can be estimated with β as dimensionless steel factor. The calculated stress has to be smaller than the maximum allowable stress σ max of the plate, given by the yield stress and the safety factor (SF). p * ( DBP) 2 0.36 σ =β* ;β = DBP 2 Tw2 1+ ( ) Hw

σ max =

σY SF

; SF ≥ 1

The part of the wall between two vertical reinforcements can be considered as a separate clamped plate on all the four sides with a uniform load p = 0,1 N/mm2. The displacement of the profiles is considered zero for the calculations of the stress and the displacement of the wall. A tank with zero reinforcements on a wall will also be calculated like a plate.

f =

0.036 p * ( DBP) 4 * ; f = f defl * DBPlim DBP 2 E * Tw3 ) Hw

1+ (

1

DBPlim 2  4  1+ ( )   f defl * DBPlim * E * Tw3 Hw  * DBPdefl =    0.036 p     0.36 p * ( DBP) 2 * σ = 2 DBP 2 Tw 1+ ( ) Hw

σ max * Tw2

DBPstress =

p

DBP 2 ) Hw 0.36

1+ ( *

If one of the values of DBPdefl or DBPstress is smaller than the value of DBPlim, the maximum distance between profiles is limited by the deflection or stress. In this cased the value of DBPlim is equal to the smallest value of these two. DBPdefl < DBPlim ⇒ DBPlim = DBPdefl If DBPstress < DBPlim ⇒ DBPlim = DBPstress

Maximum Plate dimensions

Distance Between Profiles

The maximum values of the plate dimensions are limited by the maximum allowable stress. The dimensions are also limited by the maximum deflection, but in order to get an starting value of these dimensions an iterative process starts with the maximum stress. Afterwards these values have to be checked for the maxip * ( DBP) 2 0.36 * σ max = DBP 2 Tw2 1+ ( ) Hw

When the maximum Distance Between Profiles (DBPmax) is fixed, according to the previous paragraph, the real DBP can be calculated. This DBP depends of the number of reinforcements and the width of the profile.

DBP 2 ) Hw ; DBP DBP then the program will place the last reinforcement outside the dimensions of the tank. In that case the position of the Lifting Lugs have to be calculated by hand.

5. Computer program With the sizes of the input files, given during the research, theoretically H Profile 7 U Profile 300 I Profile 16 L Profile 42

4 336 14400 768 2016 17520

During the building of the program it occured several times that the internal memory of Turbo pascal was not large enough to run the program in the correct way. This problem was caused by the large number of solutions created. To solve this problem the number of memorized solutions had to be diminished. This is done in several ways troughout the whole program: • During the combination of the different solutions for the short and the long wall, for the long wall only the best combination with a solution for a short wall is memorized, because it is assumed that the less weight solution will be chosen. • For each side of the tank the program will run twice. During the first run the progam memorizes of every combination of wallthickness, profiletype and number reinforcements only the solution which has the less weight. After the first run a choice is made for the type of profile. The second time the program runs only for the selected profieltype, but instead of only one memorized solution of every combination of wallthickness, profile type, and number reinforcements the first ten solutions according to the weight are memorized. So the first time the scope is wide, but there is not much depth in the solutions. While the second time the scope is small, but the depth of the solutions is larger. • Beyond a length of 5000 mm the minimum wallthickness is 8 mm. This reduces also the number of solutions. • The allowable wallthickness of the short wall is the wallthickness of the long wall +/- 2 mm.

Height of the profile (so increasing the height) the inertia per kilogram profile is Width bigger.

6. Analyse This chapter discusses the results of the assumptions and implementation of the model in the computer program.

U Profiles When the inertia and the stress of the profiles are the bottleneck for an allowable solution, the necessary increase of the inertia goes hand in hand with an increase of the total weight. Because the U profiles are popular and have very good properties for stiffen the tank wall and the U profiles can be bend in any size, an analyze is made of the relationship between the inertia and the weight of a profile. For a set of profiles with the same width (325 mm) the inertia of the profile is divided by the weight of the profile for different sizes of the height and the thickness. To be able to make a good and fare comparison between the different sizes the inertia is divided by the weight, so the graphic represents the inertia per kilogram profile.

Inertia / kg

Ratio Inertia / weight

What the optimal solution will depend on the required Inertia and stress of the profile for the sizes of the tank. But a conclusion can be made that if in the optiHeight mal solutions a profile exist with a wall thickness (8,10,12) and the ratio of Width the profile is below 0,80 then a profile has to be added to the input file U Beam with the same width, thickness 6 mm and an increase of the height. In this case an optimum value for the weight can be found. In general can be said that the minimum increase of weight for the three variables, width, height and thickness, an increase to the next step of a possible dimension, is the best for the width, then for the height and as last for the thickness of the profile.

Total Inertia The total inertia of profile and the section of the wall together, explained in the previous chapter, can be calculated as: T Itot := I p + Iw + Ap *(Yp +Tw −Yp+w)2 + Aw *(Yp+w − w )2 2 The next column chart shows the contribution of each value of the elements of the formula mentioned above for the total inertia. For a good comparison between the profiles is the inertia of the elements divided by the weight of the profile Inertia / kg per element

U profile

Accumulation of Inertia / kg

32 5x 32 14 5x 0x 32 1 6 5x 40 32 14 x8 5x 0x 1 1 32 40 0 5x x1 32 15 2 5 0 32 x1 x6 5x 50 32 15 x8 5x 0x 1 1 32 50 0 5x x1 32 16 2 5 0 32 x1 x6 5x 60 32 16 x8 5x 0x 1 1 32 60 0 5x x1 32 18 2 5 0 32 x1 x6 5x 80 32 18 x8 5x 0x 18 10 0x 12

700000 600000 500000

Trans Area Profile

400000

Inertia Profile

300000

Trans Area Wall Inertia Wall

200000 100000 0 U Profile

The graphic shows that for a profile with the same width (325 mm) and height (140 mm) but different thickness (6..12 mm ), the ratio is decreasing. This means that the inertia of the profiles is less increasing in comparison with the increase in weight. The next step in the height (150 mm) has a higher ratio than his predecessor for the same wall thickness (6..12 mm). This means that by increasing the ratio

L Profile

H Profile

I Profile

The inertia of the wall is neglect able in comparison with the other elements. The inertia by the translation of the area of the wall to the neutral axis conversely is rather a big element. Only for the I profile the inertia by the translation is small because of the small dimensions of the width of the profile,

Globally can be said that the inertia per kilogram is reduced with a factor 2/3 for respectively H, U, L, I profile.

eFunda

Displacement

Pauwels

eFunda

Stress

0.035

0.6 0.03

Dimensionless Steelfactor

The recommendation is given to add more H profiles to the input files of the programme in order to go to the optimum solution for the weight. This recommendation is valid until the inertia and the stress of the profile are no longer the bottleneck for an allowable solution.

Pauwels 0.04

Dimensionless Steelfactor

The H profile has the best total inertia per kilogram weight in comparison with the other profiles. The choice witch profile to use for stiffen the wall depends on the sizes of the wall and the required inertia and stress of the profile. The H profile is the most economic per kilogram although for small tanks it is possible that I some profiles gives enough strength and stiffness. The best solution for the example dimensions above is with U Profiles. It means that the H profiles have the largest inertia per kilogram, it can also mean that there are too strong and too heavy for the optimal solution.

0.025

0.02

0.015

0.01

0.5 0.4 0.3 0.2 0.1

0.005

After a research for the stresses and displacements in a steel plate, on the internet (eFunda) and in the Pauwels documentation two the same formulas are found. The only difference is the constant dimensionless steel factor (α and β), which the stresses and displacements are multiplied. p * ( DBP) 4 p * ( DBP) 2 f =α * ; σ =β* E * Tw3 Tw2 The displacement and stress is plotted in the graphics below, the only difference is the constant dimensionless steel factor. The displacement shows a little difference between the two, the stress however shows a big difference and more than 50 %. The values used with eFunda are discreet and determined by a least Square Method (Appendix xx

Ratio Height/DBP

Ratio Height/DBP

Maximum Distance Between Profiles If the distance between the profiles is considered as a clamped plate, the maximum dimension of this plate is limited by the deflection and the stress. T σ max Since DBPlim ≅ w * and σmax and p are known values, the DBP only depends p 0 .6 on the wall thickness and therefore: DBPlim ≅ 80 * Tw

1

DBPlim 2  4  1+ ( )   f defl * DBPlim * E * Tw3 Hw   DBPdefl = *   p 0.036     With DBPlim=80*Tw and DBPlim Constitutive -> Resultants -> Equilibrium =

Plate Equation

which yields

Displacement Note that homogeneous material across the plate (x and y directions) is assumed. As a final step, assuming homogeneous material along the thickness of the plate, the bending stiffness of the plate can be written as

where values of c1 are listed in the following table.

We then arrive at the Classical Plate equation,

or a slimmer form

Max(Lx/Ly, Ly/Lx) 1.0 1.2 1.4 1.6 1.8 2.0 c1 0.0138 0.0188 0.0226 0.0251 0.0267 0.0277 0.0284

The formula is valid for most commonly used metal materials that have Poission's ratios around 0.3. In fact, the Poisson's ratio has a very limited effect on the displacement and the above calculation normally gives a very good approximation for most practical cases. The coefficient c1 is calculated by the polynomial least-square curve-fitting.

Stress where w0 is replaced by w and pz replaced by p to be consistent with the notations in most published literatures.

where values of c2 are listed in the following table. Max(Lx/Ly, Ly/Lx) 1.0 1.2 1.4 1.6 1.8 2.0 c2 0.3078 0.3834 0.4356 0.4680 0.4872 0.4974 0.5000 The formula is valid for most commonly used metal materials that have Poission's ratios around 0.3. The coefficient c2 is calculated by the polynomial least-square curve-fitting.

Input

Flowchart main program

Initialize

Side=2 Assign side

Side=1

Assign wall thickness

Calculate if zero reinforcements on a wall is possible Next Wall thickCalculate maximum DBP and cornerdistance left and right

Assign profile type

Read and memorizes data of one profile type

Assign number of reinforcements Next profile

Find a solution Number reinforcements First Sorting

Maximum number of reinforcements No

Yes

Calculate inertia and area of Last profile type

Output

Flowchart function solution Input

No

Yes Maximum wall thickness

Yes

Initialize

No

Read profile size data

Sorting the found solutions

Next profile size Calculate Distance between profiles

Second run of program for given side

No

No

Yes

(DBP>30*Wall Thickness) and (DBPIrequired Yes Calculate frame dimensions

No

Calculate required moment of resistance Calculate total tank weight

Locate lifting lugs on right reinforcement

Sprofile>Srequired Yes

No

9. Appendix Profiles

Calculate deflection of wall

H Profiles fwall